Cow Exhibition
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10882 | Accepted: 4309 |
Description
"Fat and docile, big and dumb, they look so stupid, they aren‘t much
fun..."
- Cows with Guns by Dana Lyons
The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow.
Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si‘s and, likewise, the total funness TF of the group is the sum of the Fi‘s. Bessie wants to maximize the sum of TS and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative.
Input
* Line 1: A single integer N, the number of cows
* Lines 2..N+1: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow.
Output
* Line 1: One integer: the optimal sum of TS and TF such that both TS and TF are non-negative. If no subset of the cows has non-negative TS and non- negative TF, print 0.
Sample Input
5 -5 7 8 -6 6 -3 2 1 -8 -5
Sample Output
8
Hint
OUTPUT DETAILS:
Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF
= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the value
of TS+TF to 10, but the new value of TF would be negative, so it is not
allowed.
很烦的题目，想了一个晚上，看了大牛的博客，貌似越懂越懂。。
大牛博客解释：
今天遇到一题poj2184，大概思路是01背包dp之后把符合要求的最优解统计出来。但是在解01背包的时候遇到一个问题是体积有负数，这样在dp的过程中会遇到两个问题：循环的时候超出体积的范围；压缩空间的时候状态转移方程：dp[v]=max(dp[v],dp[v-c[i]]+w[i])，c[i]为负数时v-c[i]>v，这样按一般的循环的方向从大到下会重复计算。
先看第二个问题，在一般的01背包压缩空间的时候，体积的遍历是从大到小，因为dp[v]=max(dp[v],dp[v-c[i]]+w[i])，当前的dp[v]只取决于比自己小的dp[v-c[i]]，所以从大到小遍历时每次dp[v-c[i]]和dp[v]都是上一次的状态。
如果体积为负v-c[i]>v，从大到小遍历dp[v-c[i]]是当前物品的状态，不是上一个，这样就会出错，解决的办法是从小到大遍历。
针对第一个问题，在处理的时候将整个数轴平移，使得原来所有可能的情况都为正。
例如这题，首先计算出数据的范围：
一共100组数，从-1000到1000，那么体积的范围就是-100*1000到100*1000。平移之后我们要处理的数据范围就在0到200000，新的原点变成100000。初始化变成：
题意：题目要求选出一些牛，使smartness和funness值的和最大，而这些牛有些smartness或funness的值是负的，还要求最终的smartness之和以及funness之和不能为负。
附上代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5 #define S 100000
6 #define M 200000
7 #define INF 0x3f3f3f3f
8 int dp[M+5];
9 int max(int a,int b)
10 {
11 return a>b?a:b;
12 }
13 int main()
14 {
15 int n,m,i,j;
16 int a[105],b[105];
17 while(~scanf("%d",&n))
18 {
19 for(i=0; i<n; i++)
20 scanf("%d%d",&a[i],&b[i]);
21 memset(dp,-INF,sizeof(dp)); //初始化为极小值
22 dp[S]=0; //100000代替0，去除负数
23 for(i=0; i<n; i++)
24 {
25 if(a[i]>0) //大于0,从右到左常规推导
26 {
27 for(j=M; j>=a[i]; j--)
28 if(dp[j-a[i]]>-INF)
29 dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
30 }
31 else //小于0，则需要从左到右推导
32 {
33 for(j=0; j-a[i]<=M; j++)
34 if(dp[j-a[i]]>-INF)
35 dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
36 }
37 }
38 int sum=0;
39 for(i=S; i<=M; i++)
40 if(dp[i]>=0) //都必须大于0
41 sum=max(sum,dp[i]+i-S);
42 printf("%d\n",sum);
43 }
44 return 0;
45 }